User blog:Mh314159/A new approach
After two attempts to climb up the FGH and not getting very far, here's another approach. Hopefully I have learned something from my first two tries. Please note that there may be typography problems in this thing relating to nested subcripts and superscripts. Thank for reading! At first, if you don't want to read the whole thing, please just read down to where I recurse 1,2 and please verify that it terminates! Even a simple expression like this was a challenge for me to visualize. fn(1) = f0(n+1) f0(x) = x + 1 fn(x) = fn-1fn(x-1)(x) within brackets, always drop zeroes and collapse a = fa(a) 1 = f1(1) = f0(2) = 3 f1(2) = f0f1(1)(2) = f03(2) = 5 f1(3) = f05(3) = 8 f1(4) = f08(4) = 12 f1(x) = 3 + (x2 + x - 2)\2 2 = f2(2) = f1f2(1)(1) = f14(1) = f13(3) = f12(8) = f1(38) = 728 3 = f3(3) = f2f3(2)(3) Now I will define strings in the brackets and recurse them using f a,b = f[a-1,b,b-1](a) a,b,c = f[a-1,b,c,a,b-1,c,c-1](a) etc. 1,2 = f[2,1](1) = f728,1(1) where 728,1 = f727,1(728) etc. where 727,1 = f726,1(727) and eventually the subscript recurses to 1,1 where 1,1 = f1(1) = f3(1) = f0(1+3) = 2+3 Now I will define very long strings in the brackets using rows of brackets. The next function after f will recurse these rows. x is any number or string y is a string n is a number x0 = x xn = p,p,...,pn-1 where p = xn-1 and there are p instances of p xy = xn where n = y 11 = 3,3,3 21 = 728,728,...,728 and there are 728 entries. 22 = p,p,...,p where p = 21 and there are p entries. 22,1 = 2n where n is the value of 2,1 if n is a number eliminate 0s and collapse xyn = x [xn-1n] n-1 (notice that number of brackets remains the same until the final bracket decrements to zero) 2,231 = 2,2 [2,221] 0 = 2,2[2,221] 2,221 = 2,2 [2,211] 2,211 = 2,2 [2,21] Therefore, 2,231 = 2,2[2,2[2,2[2,21]]] This means first find 2,21 and then use that as the recursion number m in 2,2m and then use 2,2m as the recursion number p in 2,2p and then use 2,2p as the recursion number r in 2,2r in general, if n is a number, xy...zn (s sets of brackets) = xy...[xy...z-1n] n-1 (s sets of brackets) if q is a string, xy...q = xy...n where n = the value of q 4321 = 43[4311]0 = 43[4311] where 4311 = 43[4301]0 = 43[431] Now the next function number. The symbol group f{N} indicates a function and should be treated as if it was simply an f or a g. f{1}n(1) = f{1}0(n+1) f{1}0(x) = x,x,...,x...x,x,...,x with x x's in each string and x sets of brackets f{1}n(x) = f{1}n-1f{1}n(x-1)(x) a1 = f{1}a(a) 11 = f{1}1(1) = f{1}0(2) = 2,2,...,2...2,2,...,2 and because 2 = 728, there are 728 entries in each string and 728 sets of brackets f{1}2(1) = f{1}0(3) f{1}1(2) = f{1}0f{1}1(1)(2) = f{1}0f{1}0(2)(2) 21 = f{1}2(2) = f{1}1f{1}2(1)(2) = f{1}1f{1}0(3)(2) = f{1}1(f{1}1...f{1}1(2)) with f{1}0(3) recursions a,b1 = f{1}[a-1,b1,b-1]1(a) a,b,c1 = f{1}[a-1,b,c1,a,b-1,c1,c-1]1(a) etc. 1,11 = f{1}[11]1(1) = f{1}n1(1) where n = 11 (a large number, see above) and the subscript n1 = f{1}n(n) x10 = x1 x1n = p,p,...,p1n-1 where p = x1n-1 and there are p instances of p x1y = x1n where n = y x10 = x1 x1n1 = x1r r = the value of n1 x1y1 = x1n where n = the value of y1 211 = p,p,...,p1 with p = 21 and with p entries. 212 = p,p,...,p1 with p = 211 and with p entries. 2111 = 21r with r = 11 2122 = 21r with r = 22 Generalizing to any function number m f{m}n(1) = f{m}0(n+1) f{m}0(x) = x,x,...,xm-1...x,x,...,xm-1 with xm-1 entries in each string and xN-1 sets of brackets f{m}n(x) = f{m}n-1f{m}n(x-1)(x) am = f{m}a(a) a,bm = f{m}[a-1,bm,b-1]m(a) a,b,cm = f{m}[a-1,b,cm,a,b-1,cm,c-1]m(a) etc. xm0 = xm xmn = p,p,...,pmn-1 where p = xmn-1 and there are p instances of p xmy = xmn where n = y xmnm = xmr where r = the value of nm xmym = xmr where r = the value of ym Now expressions of the type 99 and even more deeply recursed subscripts are defined The stringed function number: ↓(n) indicates that n is a subscript f{0,1}n(1) = f{0,1}0(n+1) f{0,1}0(x) = x↓(x↓(...x)) with x instances of x f{0,1}n(x) = f{0,1}n-1f{0,1}n(x-1)(x) a0,1 = f{0,1}a(a) f{0,1}0(1) = 1↓(1↓1)) because 1 = 3 10,1 = f{0,1}1(1) = f{0,1}0f{0,1}0(2)(1) where f{0,1}0(2) = 2↓(2↓(...2)) with 728 instances of 2 Next to do: define strings and rows for a0,1 and then f{1,1} recurses rows of a0,1 and a1,1 = f{1,1}a(a) and repeat the process for the general function number f{m,1} Then what might f{0,2} mean? Category:Blog posts